Bài 5 nha bạn :
\(\frac{bz-cy}{a}=\frac{cx-az}{b}=\frac{ay-bx}{c}\)
\(=\frac{x\left(bz-cy\right)}{ax}=\frac{y\left(cx-az\right)}{by}=\frac{z\left(ay-bx\right)}{zc}\)
\(=\frac{bzx-cyx}{ax}=\frac{ycx-azy}{by}=\frac{ayz-bxz}{zc}\)
\(=\frac{\left(bzx-cyx\right)+\left(ycx-azy\right)+\left(ayz-bxz\right)}{ax+by+zc}\)
\(=\frac{\left(bzx-bzx\right)+\left(-cyx+ycx\right)+\left(-azy+ayz\right)}{ax+by+zc}=\frac{0}{ax+by+zc}=0\)
\(\Rightarrow\frac{bz-cy}{a}=0\Rightarrow bz-cy=0\Rightarrow bz=cy\Rightarrow\frac{z}{c}=\frac{y}{b}\) (1)
\(\Rightarrow\frac{cx-az}{b}=0\Rightarrow cx-az=0\Rightarrow cx=az\Rightarrow\frac{x}{a}=\frac{z}{c}\) (2)
\(\Rightarrow\frac{ay-bx}{c}=0\Rightarrow ay-bx=0\Rightarrow ay=bx\Rightarrow\frac{y}{b}=\frac{x}{a}\) (3)
Từ (1) ; (2) và (3) \(\Rightarrow\frac{x}{a}=\frac{y}{b}=\frac{z}{c}\) (đpcm)
Mik chỉ trả lời đc câu 3 thôi!
\(\dfrac{\left(x-5\right)^2}{7}\) =7 => \(^{\left(x-5\right)^2}\)= 49
=> \([^{x-5=7}_{x-5=-7}=>[^{x=12}_{x=-2}\)
p mình với.